Talk:Kripke–Platek set theory
KPM > Similarly, KPM is KP asserting "the class of all ordinals is recursively Mahlo". Is there a source stating that KPM is actually equivalent to KP asserting "the class of all ordinals is recursively Mahlo"? p-adic 14:19, October 14, 2018 (UTC) :No. I got wrong definition. Now it changes into KP asserting "the universe is recursively Mahlo" (along with other large ordinal axioms). {hyp/^,cos} (talk) 00:07, October 15, 2018 (UTC) :: I see. Thank you. By the way, isn't it better to define the recursive Mahloness of a class, which is not necesarrily a set, because "the universe" is not a set? :: p-adic 03:15, October 15, 2018 (UTC) ::This definition "links" recursively Mahlo sets and recursively Mahlo ordinals. :Although it shares the name of "Mahlo cardinals", recursively Mahlo ordinals, recursively Mahlo sets, and KPM are not defined using stationary sets. Consider this "naive copy" definition: :*Ordinal \(\alpha\) is recursively Mahlo if it is admissible and \(\{\beta<\alpha|\beta\text{ is admissible}\}\) is stationary in \(\alpha\) :*Set \(x\) is recursively Mahlo if it is admissible and \(\{y\in x|y\text{ is admissible}\}\) is stationary in \(x\) :*KPM is KP asserting "the class of all admissible sets is stationary" :but the definitions didn't use them. I think there might be some points where the "naive copy" definitions go wrong. There are some alternative definitions, but not all link recursively Mahlo sets and recursively Mahlo ordinals. {hyp/^,cos} (talk) 06:09, October 15, 2018 (UTC) :: The stationary property refers to the club property, which does not make sense for a set which is not an ordinal. Therefore it is obvious that the "naive copy" definitions for a "recursively Mahlo set" and "KPM" do not work at all. Sorry if I misunderstood what you meant. :: p-adic 06:21, October 15, 2018 (UTC) KP\(\omega\) + Separation vs. ZFC\Powerset The least level of constructible hierarchy to satisfy KP\(\omega\) + Sep and ZFC\Powerset are equal. But do those two systems have equal PTO? If so, why the Choice and full Replacement in ZFC\Powerset do not improve its PTO? {hyp/^,cos} (talk) 01:54, September 30, 2019 (UTC) : Although I do not know the comparison of their PTOs, it is not so strange even if AC does not improve the PTO, is it? For example, Con(ZF) is equivalent to Con(ZFC), and PTO is partially related to Con. : p-adic 11:54, September 30, 2019 (UTC) :But what about Replacement? Replacement is the main point in the strength of ZFC, because "ZFC but the Replacement restricted to \(\Pi_n\) formulas" is weaker and has smaller PTO than ZFC. {hyp/^,cos} (talk) 13:43, September 30, 2019 (UTC) : I have no idea about the effect of Replacement here. For example, let \(\alpha\) denote the least ordinal such that \(L_{\alpha}\) is a model of a given set theory without full Replacement. Then \(\alpha\) is also the least ordinal such that \(L_{\alpha}\) is a model of the set theory augumented by Replacement for formulae which are true at \(L_{\alpha}\). The new Replacement might be very weak Replacement, and also might be the full Replacement. It depends on how good \(L_{\alpha}\) is, which depends on the set theory. : p-adic 15:31, September 30, 2019 (UTC) Collection vs. Replacement Does Collection imply Replacement? And what about \(\Sigma_n\)-Collection vs. \(\Sigma_n\)-Replacement? If so, how large (how small) will the PTO be if the \(\Sigma_0\)-Collection in KP is substituted by \(\Sigma_0\)-Replacement? {hyp/^,cos} (talk) 10:20, November 13, 2019 (UTC) : > Does Collection imply Replacement? : It is true when you assume Separation. : > And what about \(\Sigma_n\)-Collection vs. \(\Sigma_n\)-Replacement? : It is true when you assume \(\Sigma_n\)-Separation. : > how large : Sorry, I do not know. : p-adic 13:37, November 13, 2019 (UTC)